A system consits of a uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density `rho=alpha/r`, where `alpha` is a positive constant and r is the distance from the centre of the sphere. Find the charge of the sphere for which the electric field intensity E outside the sphere is independent of R.

Options

A. `alpha/(2epsilon_(0))`
B. `2/(alphaepsilon_(0))`
C. `2pialphaR^(2)`
D. None of these

Explanation

To solve the problem, we need to find the charge of a uniformly charged sphere such that the electric field intensity \( E \) outside the sphere is independent of the radius \( R \) of the sphere.

### Step-by-step Solution:

1. **Understanding the Charge Distribution**:
The charge density of the surrounding medium is given by:
\[ \rho = \frac{\alpha}{r} \]
where \( \alpha \) is a positive constant and \( r \) is the distance from the center of the sphere.

2. **Finding the Total Charge in the Medium**:
To find the total charge \( Q_{\text{medium}} \) in the surrounding medium, we need to integrate the charge density over the volume of the medium. The volume element in spherical coordinates is \( dV = r^2 \sin \theta \, dr \, d\theta \, d\phi \).

The total charge in the shell from radius \( R \) to some outer radius \( R' \) is:
\[ Q_{\text{medium}} = \int_R^{R'} \rho \, dV = \int_R^{R'} \frac{\alpha}{r} \cdot r^2 \, dr \cdot \int_0^{\pi} \sin \theta \, d\theta \cdot \int_0^{2\pi} d\phi \]
The angular integrals give \( 4\pi \), so:
\[ Q_{\text{medium}} = 4\pi \alpha \int_R^{R'} r \, dr = 4\pi \alpha \left[ \frac{r^2}{2} \right]_R^{R'} = 4\pi \alpha \left( \frac{(R')^2}{2} - \frac{R^2}{2} \right) \]

3. **Using Gauss's Law**:
According to Gauss's law, the electric flux through a closed surface is equal to the charge enclosed divided by \( \epsilon_0 \):
\[ \Phi_E = \oint E \cdot dA = \frac{Q_{\text{enc}}}{\epsilon_0} \]
For a Gaussian surface of radius \( r > R' \), the total charge enclosed \( Q_{\text{enc}} \) is the sum of the charge of the sphere \( Q \) and the charge in the medium:
\[ Q_{\text{enc}} = Q + Q_{\text{medium}} \]

4. **Setting Up the Equation**:
The electric field \( E \) at a distance \( r \) from the center of the sphere is given by:
\[ E \cdot 4\pi r^2 = \frac{Q + Q_{\text{medium}}}{\epsilon_0} \]
Therefore:
\[ E = \frac{Q + Q_{\text{medium}}}{4\pi \epsilon_0 r^2} \]

5. **Ensuring Independence from R**:
For \( E \) to be independent of \( R \), the term \( Q + Q_{\text{medium}} \) must not depend on \( R \). From our expression for \( Q_{\text{medium}} \):
\[ Q_{\text{medium}} = 4\pi \alpha \left( \frac{(R')^2}{2} - \frac{R^2}{2} \right) \]
We can simplify this to:
\[ Q_{\text{medium}} = 2\pi \alpha \left( (R')^2 - R^2 \right) \]
To ensure that \( E \) is independent of \( R \), we set:
\[ Q + 2\pi \alpha (R^2) = 0 \]
This implies:
\[ Q = -2\pi \alpha R^2 \]

6. **Conclusion**:
Thus, the charge of the sphere for which the electric field intensity \( E \) outside the sphere is independent of \( R \) is:
\[ Q = -2\pi \alpha R^2 \]